| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Independence test requiring preliminary calculations |
| Difficulty | Moderate -0.8 This is a straightforward application of standard independence testing and conditional probability formulas. Part (i) requires checking if P(W∩C) = P(W)×P(C), part (ii) is basic Venn diagram completion using given probabilities, part (iii) applies the definition P(W|C) = P(W∩C)/P(C), and part (iv) requires simple comparison of two probabilities. All calculations are routine with no problem-solving insight needed, making this easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
2 In the 2001 census, people living in Wales were asked whether or not they could speak Welsh. A resident of Wales is selected at random.
\begin{itemize}
\item $W$ is the event that this person speaks Welsh.
\item $C$ is the event that this person is a child.
\end{itemize}
You are given that $\mathrm { P } ( W ) = 0.20 , \mathrm { P } ( C ) = 0.17$ and $\mathrm { P } ( W \cap C ) = 0.06$.\\
(i) Determine whether the events $W$ and $C$ are independent.\\
(ii) Draw a Venn diagram, showing the events $W$ and $C$, and fill in the probability corresponding to each region of your diagram.\\
(iii) Find $\mathrm { P } ( W \mid C )$.\\
(iv) Given that $\mathrm { P } \left( W \mid C ^ { \prime } \right) = 0.169$, use this information and your answer to part (iii) to comment very briefly on how the ability to speak Welsh differs between children and adults.
\hfill \mbox{\textit{OCR MEI S1 2008 Q2 [8]}}